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A metric space is a complete metric space when every Cauchy sequence converges to a point of that space: for each $\epsilon >0$, the condition $d(x_m,x_n)<\epsilon$ for all sufficiently large $m,n$ must imply existence of $x$ with $d(x_n,x)\to 0$.

A Cauchy sequence in $[0,1]$ is also Cauchy in $\mathbb{R}$, so it converges there by the assumed completeness. Its limit remains between zero and one, since inequalities are preserved under limits. ThereforeThis is an instance of the fact that a closed subset of a complete metric space is complete with its induced metric.

The Baire category theorem says that in a nonempty complete metric space, every countable intersection of open dense sets is dense. To prove it, start in an arbitrary nonempty open set $O$ and let $G_n$ be open dense. Inductively choose closed balls $\overline{B}(x_n,r_n)$ with $0<r_n\le 2^{-n}$, contained in $G_n$ and in the interior of the previous ball, with the first also contained in $O$. Density provides a point at each stage, and openness permits the smaller closed ball. The centers are a Cauchy sequence; completeness supplies a limit lying in every closed ball, hence in $O\cap {\bigcap }_n{G}_n$. The proof also works at isolated points, where sufficiently small balls are singletons.

The incomplete metric space $\mathbb{Q}$ is a counterexample: it is the countable union of its closed nowhere dense singletons, so the complements are open dense with empty intersection.

For the sequence in the problem, putEach $E_N$ is closed by continuity. Pointwise convergence makes each point's values a Cauchy sequence, so $[0,1]={\bigcup }_N{E}_N$. By the Baire category theorem, one $E_N$ has nonempty relative interior, which contains a nonempty open interval $J$ in $(0,1)$. This proves the required common tail bound.

If continuous $g_n$ converged pointwise to the rationality indicator, take $ϵ=1/4$ and let $m\to \infty$ in this bound. On $J$, $|g_n-g|\le 1/4$ for $n\ge N$. Hence a fixed continuous $g_N$ is at least $3/4$ at every rational point and at most $1/4$ at every irrational point of $J$. The density of both sets contradicts continuity. Thus the rationality indicator is not a pointwise limit of continuous functions.

Nevertheless $h_n(x)=x^n$ are continuous and converge pointwise to $h(x)=0$ for $x<1$, $h(1)=1$. This limit is discontinuous at $1$: a pointwise limit of continuous functions can be discontinuous.

One form of the Baire category theorem is: a nonempty complete metric space cannot be a countable union of closed nowhere dense subsets. Equivalently, a countable intersection of open dense subsets is dense. A Banach space is a complete metric space for its norm metric, so this form applies directly to the next part.

A complete metric space is one in which every Cauchy sequence converges. On $I=(0,1]$ defineThe map $x\mapsto 1/x$ is an isometry from $(I,d)$ onto the closed subset $[1,\infty )$ of $\mathbb{R}$, so $d$ is a metric and is complete. That map and its inverse are continuous in the ordinary Euclidean metrics, so it is also a homeomorphism. Therefore $d$ induces exactly the usual subspace topology on $I$.

One form of the Baire category theorem says that a nonempty complete metric space cannot be a countable union of closed nowhere dense sets.

For each nonnegative integer $n$, putBecause $f$ is an entire function, every derivative $f^{(n)}$ is continuous and $Z_n$ is closed. No $f^{(n)}$ is identically zero: if $f^{(n)}\equiv 0$ for some $n$, repeated integration shows that $f$ is a polynomial. By the identity theorem, the zero set of the nonzero holomorphic function $f^{(n)}$ has empty interior. Thus each $Z_n$ is nowhere dense.

The complex plane is complete, so the Baire category theorem givesChoose $z_0$ outside this union. Then $f^{(n)}(z_0)\ne 0$ for every $n$, and hence every coefficient $f^{(n)}(z_0)/n!$ in the Taylor series about $z_0$ is nonzero. Such a point $z_0$ therefore exists.

First, compactness makes $X$ totally bounded: for every positive integer $k$, the open cover by balls of radius $1/k$ has a finite subcover. Given a sequence, repeatedly choose a ball of radius $1/k$ containing infinitely many terms of the subsequence retained at the preceding stage. Taking a diagonal subsequence $(x_{n_k})$, any two terms with indices at least $k$ lie in one ball of radius $1/k$, so their distance is below $2/k$. The diagonal subsequence is therefore a Cauchy sequence.

A compact metric space is complete. To see this without assuming sequential compactness, let $(y_n)$ be Cauchy and put $F_n=\overline{\{y_m:m\ge n\}}$. These nonempty closed sets are nested and have the finite intersection property; compactness gives a point $y\in {\bigcap }_n{F}_n$. Given $\epsilon >0$, a sufficiently late tail has diameter below $\epsilon /2$, and because $y$ lies in its closure, every point of that tail lies within $\epsilon$ of $y$. Thus $y_n\to y$.

The Cauchy subsequence constructed above consequently converges in $X$. Hence every compact metric space is sequentially compact.

Let $(x_n)$ be a Cauchy sequence in a compact metric space $X$. By sequential compactness of a compact metric space, it has a convergent subsequence $x_{n_j}\to x\in X$. Given $\epsilon >0$, choose $N$ such that $d(x_m,x_n)<\epsilon /2$ whenever $m,n\ge N$, and then choose $j$ such that $n_j\ge N$ and $d(x_{n_j},x)<\epsilon /2$. For every $n\ge N$, the triangle inequality givesThus the whole sequence converges to $x$. Every Cauchy sequence converges in $X$, so $X$ is a complete metric space.

The space is not compact because it is not complete. For $n\ge 2$, define the continuous functionThese functions converge in the L1 norm to the step function $h=1_{(1/2,1]}$, because $f_n-h$ is supported on an interval of length $2/n$ and bounded in absolute value by one. They are consequently Cauchy for the metric $d$, since for sufficiently close pairs the outer minimum with $1$ does nothing.

If $(f_n)$ converged in this metric to some $f\in C[0,1]$, it would also converge to $f$ in $L^1$. Uniqueness of an $L^1$ limit would give $f=h$ almost everywhere. Continuity would then force $f=0$ on $[0,1/2)$ and $f=1$ on $(1/2,1]$, which is impossible at $1/2$. Thus this Cauchy sequence has no limit in $C[0,1]$, while every compact metric space is complete.

For each $n\ge 0$, defineEvery $f^{(n)}$ is an entire function, hence continuous, so $E_n$ is closed. The hypothesis saysThe complex plane is a complete metric space. The Baire category theorem therefore implies that some $E_N$ has nonempty interior. Thus $f^{(N)}$ vanishes on an open set, and the identity theorem giveson all of $\mathbb{C}$. If $N=0$, then $f=0$; otherwise the power series of $f$ has no term of degree $N$ or higher. Hence $f$ is a polynomial of degree at most $N-1$. This is the pointwise vanishing derivative criterion for a polynomial.

A subset $A$ of a metric space is nowhere dense when the interior of its closure is empty. One form of the Baire category theorem says that a nonempty complete metric space cannot be a countable union of nowhere dense closed sets; equivalently, a countable intersection of dense open sets is dense.

Fix $\epsilon >0$ and work in the complete interval $[1,2]$. For $N\ge 1$ define the closed setThe sets are closed by the continuity of $f$, and the assumed pointwise convergence gives $[1,2]={\bigcup }_N{E}_N$. The Baire category theorem therefore gives an $N$ and a nonempty open interval $(a,b)\subseteq [1,2]$ contained in $E_N$.

The intervals $[na,nb]$ overlap once $(n+1)a\le nb$, which holds for every sufficiently large natural number $n$. Their union consequently contains a half-line. Every sufficiently large $t$ can therefore be written as $t=nx$ with $n\ge N$ and $x\in (a,b)$, and then $|f(t)|=|f(nx)|\le \epsilon$. Since $\epsilon$ was arbitrary, $f(t)\to 0$ as $t\to \infty$.

This is always true. Uniform convergence makes $(f_n)$ a uniformly Cauchy sequence: for every $\epsilon >0$, there is $N$ such thatfor every $x\in A$ and $m,n\ge N$. Letting $x\to x_0$ through points of $A\setminus \{x_0\}$ and using the assumed limits givesThus $(y_n)$ is a Cauchy sequence. Since $Y$ is a complete metric space, $(y_n)$ converges in $Y$.

A Cauchy sequence $(x_n)$ in a metric space $(M,d)$ satisfies: for every $\epsilon >0$ there is $N$ such that $d(x_m,x_n)<\epsilon$ whenever $m,n\ge N$. A complete metric space is one in which every Cauchy sequence converges to a point of the space.

Every Cauchy sequence is bounded. Choose $N$ such that $d(x_n,x_N)<1$ for $n\ge N$, and putThen every term lies in the ball $B(x_N,R)$.

Now suppose $M$ is complete and $(F_n)$ is a decreasing sequence of nonempty closed sets with $\mathrm{diam}{F}_n\to 0$. Choose $x_n\in F_n$. Given $\epsilon >0$, choose $N$ with $\mathrm{diam}{F}_N<\epsilon$. For $m,n\ge N$, both points lie in $F_N$, so $d(x_m,x_n)<\epsilon$. Completeness gives $x_n\to x\in M$. For each fixed $N$, the tail lies in the closed set $F_N$, hence $x\in F_N$. Therefore

Conversely, assume the nested-set property and let $(x_n)$ be Cauchy. DefineThese sets are nonempty, closed, and decreasing. The Cauchy property implies $\mathrm{diam}{F}_n\to 0$; taking a closure does not change the diameter. Choose $x\in {\bigcap }_n{F}_n$. Since $x_n,x\in F_n$,Thus every Cauchy sequence converges and $M$ is complete. This proves the Cantor intersection theorem characterization.

The contraction mapping theorem states that a contraction of a nonempty complete metric space has a unique fixed point.

For each $\lambda \in \Lambda$, the map $T_{\lambda }(x)=T(\lambda ,x)$ is a contraction with the common constant $k$, so it has a unique fixed point $x^{\ast }(\lambda )$. This defines the required unique function. Fix ${\lambda }_0$. The fixed-point identities and the triangle inequality giveConsequentlyThe numerator tends to zero as $\lambda \to {\lambda }_0$ by the assumed continuity for the fixed point $x^{\ast }({\lambda }_0)$. Hence $x^{\ast }$ is continuous, an instance of continuous dependence of the fixed point of a uniform contraction.

The Baire category theorem states that every countable intersection of open dense subsets of a complete metric space is dense. Equivalently, no nonempty open subset of a complete metric space is a countable union of nowhere dense sets.

To prove the first form, let $G_1,G_2,\dots$ be open dense subsets of a complete metric space $X$, and let $V\subseteq X$ be nonempty and open. Choose a closed ballInductively, density and openness of $G_{n+1}$ allow a closed ballThe balls are nested and $d(x_m,x_n)<2^{-n}$ for $m>n$, so $(x_n)$ is Cauchy. Completeness gives $x_n\to x\in X$. For every $n$, all later centres lie in $\overline{B}(x_n,r_n)$; closedness gives $x\in \overline{B}(x_n,r_n)\subseteq G_n$. The first ball also lies in $V$, henceSince every nonempty open $V$ meets the intersection, that intersection is dense.

For $n\ge 0$, putEach $S_n$ is closed, the hypothesis gives $\mathbb{R}={\bigcup }_n{S}_n$, and $E_n\subset E_{n+1}$ because a function that vanishes on an open set has derivative zero there.

The open setis dense. Indeed, inside any nonempty open interval choose a nondegenerate closed subinterval $J$. The complete metric space $J$ is covered by the closed sets $J\cap S_n$, so the Baire's theorem makes one of them contain a relative open interval, which lies in some $E_n$.

On each connected component $I$ of $\Omega$, the function $F$ is one polynomial. To see this, any compact subinterval $K\subset I$ is covered by the increasing family $(E_n)$. A finite subcover therefore gives $K\subset E_N$ for some $N$, and $F^{(N)}=0$ on $K$. Hence $F$ is polynomial on $K$; overlapping compact intervals force these polynomials to agree throughout $I$.

Let $D=\mathbb{R}\setminus \Omega$. This closed set has no isolated points. Otherwise, on the two sides of an isolated point $x$, $F$ would equal polynomials. Smoothness makes all their one-sided derivatives agree at $x$, so the two polynomials are identical and extend across $x$. Some derivative would then vanish on a neighborhood of $x$, contrary to $x\in D$.

Suppose that $D$ is nonempty. It is complete andis a countable closed cover. A second application of Baire's theorem gives an index $k$ and an open interval $U_0$ such thatChoose $x_0\in D\cap U_0$ and a bounded open interval $U$ centred at $x_0$ whose closure lies in $U_0$.

Because $D$ has no isolated points, every $x\in D\cap U$ is approached by distinct points of $D\cap U$. Difference quotients first give $F^{(k+1)}(x)=0$, and induction gives

No component of $\Omega$ meeting $U$ can cross $x_0$, so it has an endpoint in $D\cap U$. If its polynomial had degree $d\ge k$, then its $d$th derivative would approach a nonzero constant at that endpoint, contradicting the preceding display. Its degree is therefore less than $k$, so $F^{(k)}$ also vanishes on the part of that component in $U$. We conclude that $F^{(k)}=0$ throughout $U$, which says $U\subset E_k\subset \Omega$ and contradicts $D\cap U\ne \varnothing$.

The pointwise inequality $\hat{g}(X,X)\ge g(X,X)$ impliesEvery $d_{\hat{g}}$-Cauchy sequence $(x_j)$ is therefore $d_g$-Cauchy. Since $g$ is geodesically complete, the Hopf-Rinow theorem makes $(M,d_g)$ a complete metric space, so $x_j\to x$ in $d_g$ for some $x\in M$.

On a coordinate neighbourhood with compact closure around $x$, smooth positive-definite Riemannian metrics are uniformly equivalent. Thus there is $C>0$ such thatthere. For all sufficiently large $j$, a short $g$-geodesic from $x$ to $x_j$ stays in this neighbourhood, and henceThus $(M,d_{\hat{g}})$ is complete. Another application of Hopf-Rinow shows that $\hat{g}$ is geodesically complete.